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lecture17 - Friction I o W Friction I C 0 If at rest 0:21...

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Unformatted text preview: Friction I o W Friction I C 0 If at rest, 0:21? 3’ FBD of Block at rest . on amugh horizontal surface As P increases so does F so long as F S p -N S F F max Value of P for Impending Motion No Motion Friction I tan(9)=£ N -W-sin(Q) +F=0 1/ ——W-cos(6) + N=O (- i '\ \ x \ At impending motion F: us- N /7 o F. . '0 o rlctlon Probleml Frictionless pulley / 2 p s - .2 W== 50- lb for impending motion @ v.49 Pb“? Weight of G=? 20 deg 0 Friction I o o A 20 deg 4‘" a) (”fix CW-sin(20-deg)+q— F=0 V 1) Problem 1 Q\O =EF»y —w-cps(20-deg) +N=0 ‘ 2) $20 deg N N=W-cos(20- deg ) N=46.985\1b or Impendmg mono“ F ”s‘ N §Subst1tute F into 1) :p .N=.2.(‘46.985) . s G=F+W-sm(20-deg) F=9.3971b G=9.397+50-sin(20-deg) G=26.4981b ‘ ~- Friction I Frictionless pulley 2 ps- .2 for impending motion up the slope Recall that answer Weight of G= 26.498 lb \_ 6.3 20 deg Friction I F rictionless pulleys Problem7 in Class - SpL-vuvg [.2le ‘FGQZ emukT- Gnu/ES N = 5‘0, 33“). F \ 'T": Carp/Lb/V’: cf 0 f o The weights of blocks A and B are WA = 1000 lb and WB = 100 lb. If the coefficient of friction is 0.30 at all surfaces, determine the force P required to produce impending motion of block A. Fflom {3%9 0-5— B FQOM FED .OF‘A. @ZF’k: O \ @@§FU5 = O / / N13." N3Ge-cbl‘3‘ +E3¢M71=o N FIOOOH’ F2: 0 girsws 7‘— k “(BIN x: O “’0 C N_ /o 00 b 1.) @7513— N393 ~5_/)+.'3N (.309): Ftl'N‘Z.=O/ N2: 125,15 @‘QW‘WW , — Sakv‘fi' ®+® Sinukrspyausky \/ @z F = ”1.6:? '7! kl; 5 o N =?\ I ~ Q], a 75 217%; N,” ’7/7‘3‘4: —P+-Fa"‘F Mfi1+N3&%I?030 -mo ._ p +. 3(175 9.3 + .301! 2.9 1085+ 429.1155, _JV 0\ Friction I 0 0 Problem 4 in Class Determine value of 2 required to start the blocks moving -,_-P A free-body diagram for block B is shown at the right. WE = mBg + 100(9.807) = 980.7 N For impending motion: T + :Fy = p sin 25° + Fn - 980.7 = o —9 + ZF P cos 25° - F — T x f p cos 25° — 0.10Fn - 287.7 = 0 Solving yields: P = 406.7 N E 407 N Ans. o" . . O O Frletlonl Problem .. ‘25 deg in Lass mA=50'kg Determine value of 2 required to start the blocks mB=100'kg moving QT SOLUTION 'A free-body diagram for bloek A is shown at the right. WA = mAg350(9.807) = 490.4 N For impending motion: F — 490.4 cos 30° = 0 F = 424.7 N Y n n + .1 / [‘1 '7] II + \u M P11 K II T — Ff — W sin 300 r — o.10r424.7) - 490.4 sin 30° = 0 T 287.7 N I I! Friction I o o Problem 3 in Class For both surfaces Ll S = .25 Determine P required to cause impending motion 380 Q o 0 Friction I Problem 3 in Class WB=75-(9.8)=735.53N 4x. 0::Fb %[email protected]=ZF$Z 1.39..“ WB— An P ~ A f— Bf=0 B , 735.5; 392.27=0 . . n Impending Motlon BI: 735-53* 39227:“ P - .25-(39227) — .25-(1127.80.=0 0 Friction I WA: 50(9.807)=4__90.35 N fl» 0=EFX ., A“? T-cos(45-deg )=0 Impending Motion Af=g5~An 1 \ i _ = , ,) n 25(131 _'1)os(45 deg) 0 L! An + T-sin(45-cieg) — WA=O ‘ N 2) 1A+T-lgin(45-de’g)a.49q.35§=0 M” 5:1 .4 Solving Eq 1)& 2) Gives: An=392.27N T=138.692N ~ \ F BD of block A 0 Problem 3 in Class If 0 Friction I C O SQON WA: 1900“,! COEF‘: 0F Frau-mm: AQ‘B : .3 \ WFHOQON ”Ala—”3“ do; : BElEKW‘Ifl/CO 6;— ol M ‘/~_—..-xm “a? . (L) K53 THE“ max F0 wage"? +01 WNCH , ’ Txfe <5?er :5 IN Pautufiwwm " ®£fi= o - Sac NQamK3o°3+FA= o —1000N+$00Nm;n-$o°=° "1 NA: {/75ij “A? 40m No sm0© Put—D, /1. FM: ‘759-N.9‘3> , $INCE FA< PM. 8: EW-é SEES-Z Nd» ‘:>¢,0° No SkIPCQ Alf-8' Friction -Hoood— I750 +NB= 0 NB: §7SOMM aka-75 om ,= 14,512; K‘ o ’S‘MP WtkL, occun pv- B mm Fgoon‘ (NOTEQULLIg) _ m m _. m _ m _. _ m m m m m m m m m m r _ m _. Impending Motion By Sliding P >- F,=/fi2§/_ The ;00—lb crate is being moved by a rope thagasses over a smooth pulley. The COeffiQE of friction between the crate and the floor is 0.30. a. Assume that h = 4 ft and determine the force P necessary to produce impending motion. b. Determine the value of h for which impending motion by slipping and by tipping would occur simultaneously. V 0L) CA9? ]: NOTKFFING. oNk‘f $kl? @Bo-r'ron, (k—zfr———a§ ? @ZF ——-— o 5 ”3 1:7 N—‘Lbckg‘ u N=aookb 8MP ON l‘aeTTOFy Fan N r-eho P -.-. (,0qu Skle’m/O 000 IM P511; 9mg {Hwy—(am WiLL occa,‘ w HFW O- O b) Fort. T’IPPWG 36% \PPer 1-5:? F = QBX'L o a» QOUQ“C3~00 =0 W2=5O lb W1 =20 lb All Coef friction = .2 ...
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